3.422 \(\int e^{-2 \coth ^{-1}(a x)} (c-\frac {c}{a x})^3 \, dx\)

Optimal. Leaf size=54 \[ -\frac {c^3}{2 a^3 x^2}+\frac {5 c^3}{a^2 x}+\frac {11 c^3 \log (x)}{a}-\frac {16 c^3 \log (a x+1)}{a}+c^3 x \]

[Out]

-1/2*c^3/a^3/x^2+5*c^3/a^2/x+c^3*x+11*c^3*ln(x)/a-16*c^3*ln(a*x+1)/a

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Rubi [A]  time = 0.14, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {6167, 6131, 6129, 88} \[ -\frac {c^3}{2 a^3 x^2}+\frac {5 c^3}{a^2 x}+\frac {11 c^3 \log (x)}{a}-\frac {16 c^3 \log (a x+1)}{a}+c^3 x \]

Antiderivative was successfully verified.

[In]

Int[(c - c/(a*x))^3/E^(2*ArcCoth[a*x]),x]

[Out]

-c^3/(2*a^3*x^2) + (5*c^3)/(a^2*x) + c^3*x + (11*c^3*Log[x])/a - (16*c^3*Log[1 + a*x])/a

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 6129

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)*(x_))^(p_.), x_Symbol] :> Dist[c^p, Int[(u*(1 + (d*x)/c)
^p*(1 + a*x)^(n/2))/(1 - a*x)^(n/2), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[a^2*c^2 - d^2, 0] && (IntegerQ
[p] || GtQ[c, 0])

Rule 6131

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_))^(p_.), x_Symbol] :> Dist[d^p, Int[(u*(1 + (c*x)/d)
^p*E^(n*ArcTanh[a*x]))/x^p, x], x] /; FreeQ[{a, c, d, n}, x] && EqQ[c^2 - a^2*d^2, 0] && IntegerQ[p]

Rule 6167

Int[E^(ArcCoth[(a_.)*(x_)]*(n_))*(u_.), x_Symbol] :> Dist[(-1)^(n/2), Int[u*E^(n*ArcTanh[a*x]), x], x] /; Free
Q[a, x] && IntegerQ[n/2]

Rubi steps

\begin {align*} \int e^{-2 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^3 \, dx &=-\int e^{-2 \tanh ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^3 \, dx\\ &=\frac {c^3 \int \frac {e^{-2 \tanh ^{-1}(a x)} (1-a x)^3}{x^3} \, dx}{a^3}\\ &=\frac {c^3 \int \frac {(1-a x)^4}{x^3 (1+a x)} \, dx}{a^3}\\ &=\frac {c^3 \int \left (a^3+\frac {1}{x^3}-\frac {5 a}{x^2}+\frac {11 a^2}{x}-\frac {16 a^3}{1+a x}\right ) \, dx}{a^3}\\ &=-\frac {c^3}{2 a^3 x^2}+\frac {5 c^3}{a^2 x}+c^3 x+\frac {11 c^3 \log (x)}{a}-\frac {16 c^3 \log (1+a x)}{a}\\ \end {align*}

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Mathematica [A]  time = 0.15, size = 56, normalized size = 1.04 \[ -\frac {c^3}{2 a^3 x^2}+\frac {5 c^3}{a^2 x}+\frac {11 c^3 \log (a x)}{a}-\frac {16 c^3 \log (a x+1)}{a}+c^3 x \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(c - c/(a*x))^3/E^(2*ArcCoth[a*x]),x]

[Out]

-1/2*c^3/(a^3*x^2) + (5*c^3)/(a^2*x) + c^3*x + (11*c^3*Log[a*x])/a - (16*c^3*Log[1 + a*x])/a

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fricas [A]  time = 0.49, size = 62, normalized size = 1.15 \[ \frac {2 \, a^{3} c^{3} x^{3} - 32 \, a^{2} c^{3} x^{2} \log \left (a x + 1\right ) + 22 \, a^{2} c^{3} x^{2} \log \relax (x) + 10 \, a c^{3} x - c^{3}}{2 \, a^{3} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c/a/x)^3*(a*x-1)/(a*x+1),x, algorithm="fricas")

[Out]

1/2*(2*a^3*c^3*x^3 - 32*a^2*c^3*x^2*log(a*x + 1) + 22*a^2*c^3*x^2*log(x) + 10*a*c^3*x - c^3)/(a^3*x^2)

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giac [A]  time = 0.14, size = 53, normalized size = 0.98 \[ c^{3} x - \frac {16 \, c^{3} \log \left ({\left | a x + 1 \right |}\right )}{a} + \frac {11 \, c^{3} \log \left ({\left | x \right |}\right )}{a} + \frac {10 \, a c^{3} x - c^{3}}{2 \, a^{3} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c/a/x)^3*(a*x-1)/(a*x+1),x, algorithm="giac")

[Out]

c^3*x - 16*c^3*log(abs(a*x + 1))/a + 11*c^3*log(abs(x))/a + 1/2*(10*a*c^3*x - c^3)/(a^3*x^2)

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maple [A]  time = 0.04, size = 53, normalized size = 0.98 \[ -\frac {c^{3}}{2 x^{2} a^{3}}+\frac {5 c^{3}}{a^{2} x}+c^{3} x +\frac {11 c^{3} \ln \relax (x )}{a}-\frac {16 c^{3} \ln \left (a x +1\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c-c/a/x)^3/(a*x+1)*(a*x-1),x)

[Out]

-1/2*c^3/x^2/a^3+5*c^3/a^2/x+c^3*x+11*c^3*ln(x)/a-16*c^3*ln(a*x+1)/a

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maxima [A]  time = 0.31, size = 51, normalized size = 0.94 \[ c^{3} x - \frac {16 \, c^{3} \log \left (a x + 1\right )}{a} + \frac {11 \, c^{3} \log \relax (x)}{a} + \frac {10 \, a c^{3} x - c^{3}}{2 \, a^{3} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c/a/x)^3*(a*x-1)/(a*x+1),x, algorithm="maxima")

[Out]

c^3*x - 16*c^3*log(a*x + 1)/a + 11*c^3*log(x)/a + 1/2*(10*a*c^3*x - c^3)/(a^3*x^2)

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mupad [B]  time = 0.08, size = 51, normalized size = 0.94 \[ c^3\,x-\frac {\frac {c^3}{2}-5\,a\,c^3\,x}{a^3\,x^2}+\frac {11\,c^3\,\ln \relax (x)}{a}-\frac {16\,c^3\,\ln \left (a\,x+1\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((c - c/(a*x))^3*(a*x - 1))/(a*x + 1),x)

[Out]

c^3*x - (c^3/2 - 5*a*c^3*x)/(a^3*x^2) + (11*c^3*log(x))/a - (16*c^3*log(a*x + 1))/a

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sympy [A]  time = 0.35, size = 42, normalized size = 0.78 \[ c^{3} x + \frac {c^{3} \left (11 \log {\relax (x )} - 16 \log {\left (x + \frac {1}{a} \right )}\right )}{a} + \frac {10 a c^{3} x - c^{3}}{2 a^{3} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c/a/x)**3*(a*x-1)/(a*x+1),x)

[Out]

c**3*x + c**3*(11*log(x) - 16*log(x + 1/a))/a + (10*a*c**3*x - c**3)/(2*a**3*x**2)

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